To find the dimensions of the ring we have 134,490,000 miles for the diameter of the orbit of Venus, which makes its circumference 422,513,784 miles in length. Then dividing the volume of the ring, i.e. 5,054,780,604,6516 cubic miles by this length, the area of its cross-section comes to be 11,963,821,788 square miles, which, divided by the breadth of 28,489,000 miles—that is one-half of the difference between the diameters of the Venus and Mercurian nebulæ, respectively 160,210,000 and 103,232,000 miles—makes the thickness of the ring to have been 420 miles. But the inner edge having become, in the process of separation, 3·738 times more dense than the outer one (see below) the average thickness would be reduced to 225 miles.
Mercurian Nebula.
As the volume of the nebula was 2,153,115,737,298,603,2706 cubic miles after the separation of the ring for Venus, we have to condense it into the volume of the Mercurian nebula, which at 103,232,000 miles in diameter would be 576,026,613,333,333,3336 cubic miles. Then, dividing the larger of these two volumes by the smaller, we find that the density of the Mercurian nebula must have been increased 3·738 fold over that of its predecessor. But we find the density of the Venus nebula to have been 4,465,512 times less than water, dividing which by 3·738 makes the Mercurian nebula to have been 1,194,666 times less dense than water. Dividing again this density by 773·395 we find it to have been 1545 times less than air, and 274° divided by this air density gives 0·1773463° as its absolute temperature, which corresponds to -273·8226537°.
From the Mercurian nebula 103,232,000 miles in diameter, volume of 576,026,613,333,333,3336 cubic miles, and density of 1,194,666 times less than water, we have to deduct the volume of his ring, which by [Table II]. is 92,735,000,000 cubic miles at density of water. Multiplying this volume by 1,194,666 makes the ring to have been 110,787,355,3006 cubic miles in volume at the density of the nebula, and subtracting this amount from 576,026,613,333,333,3336, we get 576,026,502,545,978,0336 cubic miles for the volume to be condensed into the nebula following.
To find the dimensions of the ring we have 71,974,000 miles for the mean diameter of the orbit of Mercury, which makes its circumference 226,113,518 miles in length. Then dividing the volume of his ring, i.e. 110,787,355,3006 cubic miles, as above, by this length, the area of its cross-section comes to be 489,963,459 square miles. Here we have to determine the breadth of the ring in a new way, that is empirically. Seeing that the breadth of the ring for the earth's system was 37,205,000 and of that for Venus 28,489,000 miles, we shall assume 20,000,000 miles for the breadth of the ring for Mercury. This will make the residuary, now the Solar nebula, to have been 31,616,000 miles in radius and 63,232,000 miles in diameter. Returning now to the area of the cross-section of the ring, that is, 489,963,459 square miles, and dividing it by the assumed breadth 20,000,000 miles, makes the thickness of the ring to have been 25 miles. But, as before, its inner edge having become 4·354 times more dense than the outer one during the process of separation (see below) the average thickness must have been only 11 miles.
Solar Nebula.
Lastly, as the volume of the nebula was
576,026,502,545,978,0336
cubic miles after the separation of the ring for Mercury, we have to condense it into the volume of the Solar nebula, which at 63,232,000 miles in diameter would be